Finite Math Examples

$f (x) = - x^{2} + 2 x + 3$

Step 1

Write $f (x) = - x^{2} + 2 x + 3$ as an equation.

$y = - x^{2} + 2 x + 3$

Step 2

Interchange the variables.

$x = - y^{2} + 2 y + 3$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $- y^{2} + 2 y + 3 = x$ .

$- y^{2} + 2 y + 3 = x$

Step 3.2

Subtract $x$ from both sides of the equation.

$- y^{2} + 2 y + 3 - x = 0$

Step 3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.4

Substitute the values $a = - 1$ , $b = 2$ , and $c = 3 - x$ into the quadratic formula and solve for $y$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 1 \cdot (3 - x))}}{2 \cdot - 1}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $2$ to the power of $2$ .

$y = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (3 - x)}}{2 \cdot - 1}$

Step 3.5.1.2

Multiply $- 4$ by $- 1$ .

$y = \frac{- 2 \pm \sqrt{4 + 4 \cdot (3 - x)}}{2 \cdot - 1}$

Step 3.5.1.3

Apply the distributive property.

$y = \frac{- 2 \pm \sqrt{4 + 4 \cdot 3 + 4 (- x)}}{2 \cdot - 1}$

Step 3.5.1.4

Multiply $4$ by $3$ .

$y = \frac{- 2 \pm \sqrt{4 + 12 + 4 (- x)}}{2 \cdot - 1}$

Step 3.5.1.5

Multiply $- 1$ by $4$ .

$y = \frac{- 2 \pm \sqrt{4 + 12 - 4 x}}{2 \cdot - 1}$

Step 3.5.1.6

Add $4$ and $12$ .

$y = \frac{- 2 \pm \sqrt{16 - 4 x}}{2 \cdot - 1}$

Step 3.5.1.7

Factor $4$ out of $16 - 4 x$ .

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Step 3.5.1.7.1

Factor $4$ out of $16$ .

$y = \frac{- 2 \pm \sqrt{4 (4) - 4 x}}{2 \cdot - 1}$

Step 3.5.1.7.2

Factor $4$ out of $- 4 x$ .

$y = \frac{- 2 \pm \sqrt{4 (4) + 4 (- x)}}{2 \cdot - 1}$

Step 3.5.1.7.3

Factor $4$ out of $4 (4) + 4 (- x)$ .

$y = \frac{- 2 \pm \sqrt{4 (4 - x)}}{2 \cdot - 1}$

Step 3.5.1.8

Rewrite $4 (4 - x)$ as $2^{2} (2^{2} - x)$ .

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Step 3.5.1.8.1

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm \sqrt{2^{2} (4 - x)}}{2 \cdot - 1}$

Step 3.5.1.8.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm \sqrt{2^{2} (2^{2} - x)}}{2 \cdot - 1}$

Step 3.5.1.9

Pull terms out from under the radical.

$y = \frac{- 2 \pm 2 \sqrt{2^{2} - x}}{2 \cdot - 1}$

Step 3.5.1.10

Raise $2$ to the power of $2$ .

$y = \frac{- 2 \pm 2 \sqrt{4 - x}}{2 \cdot - 1}$

Step 3.5.2

Multiply $2$ by $- 1$ .

$y = \frac{- 2 \pm 2 \sqrt{4 - x}}{- 2}$

Step 3.5.3

Simplify $\frac{- 2 \pm 2 \sqrt{4 - x}}{- 2}$ .

$y = 1 \pm \sqrt{4 - x}$

Step 3.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

Raise $2$ to the power of $2$ .

$y = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (3 - x)}}{2 \cdot - 1}$

Step 3.6.1.2

Multiply $- 4$ by $- 1$ .

$y = \frac{- 2 \pm \sqrt{4 + 4 \cdot (3 - x)}}{2 \cdot - 1}$

Step 3.6.1.3

Apply the distributive property.

$y = \frac{- 2 \pm \sqrt{4 + 4 \cdot 3 + 4 (- x)}}{2 \cdot - 1}$

Step 3.6.1.4

Multiply $4$ by $3$ .

$y = \frac{- 2 \pm \sqrt{4 + 12 + 4 (- x)}}{2 \cdot - 1}$

Step 3.6.1.5

Multiply $- 1$ by $4$ .

$y = \frac{- 2 \pm \sqrt{4 + 12 - 4 x}}{2 \cdot - 1}$

Step 3.6.1.6

Add $4$ and $12$ .

$y = \frac{- 2 \pm \sqrt{16 - 4 x}}{2 \cdot - 1}$

Step 3.6.1.7

Factor $4$ out of $16 - 4 x$ .

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Step 3.6.1.7.1

Factor $4$ out of $16$ .

$y = \frac{- 2 \pm \sqrt{4 (4) - 4 x}}{2 \cdot - 1}$

Step 3.6.1.7.2

Factor $4$ out of $- 4 x$ .

$y = \frac{- 2 \pm \sqrt{4 (4) + 4 (- x)}}{2 \cdot - 1}$

Step 3.6.1.7.3

Factor $4$ out of $4 (4) + 4 (- x)$ .

$y = \frac{- 2 \pm \sqrt{4 (4 - x)}}{2 \cdot - 1}$

Step 3.6.1.8

Rewrite $4 (4 - x)$ as $2^{2} (2^{2} - x)$ .

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Step 3.6.1.8.1

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm \sqrt{2^{2} (4 - x)}}{2 \cdot - 1}$

Step 3.6.1.8.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm \sqrt{2^{2} (2^{2} - x)}}{2 \cdot - 1}$

Step 3.6.1.9

Pull terms out from under the radical.

$y = \frac{- 2 \pm 2 \sqrt{2^{2} - x}}{2 \cdot - 1}$

Step 3.6.1.10

Raise $2$ to the power of $2$ .

$y = \frac{- 2 \pm 2 \sqrt{4 - x}}{2 \cdot - 1}$

Step 3.6.2

Multiply $2$ by $- 1$ .

$y = \frac{- 2 \pm 2 \sqrt{4 - x}}{- 2}$

Step 3.6.3

Simplify $\frac{- 2 \pm 2 \sqrt{4 - x}}{- 2}$ .

$y = 1 \pm \sqrt{4 - x}$

Step 3.6.4

Change the $\pm$ to $+$ .

$y = 1 + \sqrt{4 - x}$

Step 3.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.7.1

Simplify the numerator.

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Step 3.7.1.1

Raise $2$ to the power of $2$ .

$y = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (3 - x)}}{2 \cdot - 1}$

Step 3.7.1.2

Multiply $- 4$ by $- 1$ .

$y = \frac{- 2 \pm \sqrt{4 + 4 \cdot (3 - x)}}{2 \cdot - 1}$

Step 3.7.1.3

Apply the distributive property.

$y = \frac{- 2 \pm \sqrt{4 + 4 \cdot 3 + 4 (- x)}}{2 \cdot - 1}$

Step 3.7.1.4

Multiply $4$ by $3$ .

$y = \frac{- 2 \pm \sqrt{4 + 12 + 4 (- x)}}{2 \cdot - 1}$

Step 3.7.1.5

Multiply $- 1$ by $4$ .

$y = \frac{- 2 \pm \sqrt{4 + 12 - 4 x}}{2 \cdot - 1}$

Step 3.7.1.6

Add $4$ and $12$ .

$y = \frac{- 2 \pm \sqrt{16 - 4 x}}{2 \cdot - 1}$

Step 3.7.1.7

Factor $4$ out of $16 - 4 x$ .

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Step 3.7.1.7.1

Factor $4$ out of $16$ .

$y = \frac{- 2 \pm \sqrt{4 (4) - 4 x}}{2 \cdot - 1}$

Step 3.7.1.7.2

Factor $4$ out of $- 4 x$ .

$y = \frac{- 2 \pm \sqrt{4 (4) + 4 (- x)}}{2 \cdot - 1}$

Step 3.7.1.7.3

Factor $4$ out of $4 (4) + 4 (- x)$ .

$y = \frac{- 2 \pm \sqrt{4 (4 - x)}}{2 \cdot - 1}$

Step 3.7.1.8

Rewrite $4 (4 - x)$ as $2^{2} (2^{2} - x)$ .

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Step 3.7.1.8.1

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm \sqrt{2^{2} (4 - x)}}{2 \cdot - 1}$

Step 3.7.1.8.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm \sqrt{2^{2} (2^{2} - x)}}{2 \cdot - 1}$

Step 3.7.1.9

Pull terms out from under the radical.

$y = \frac{- 2 \pm 2 \sqrt{2^{2} - x}}{2 \cdot - 1}$

Step 3.7.1.10

Raise $2$ to the power of $2$ .

$y = \frac{- 2 \pm 2 \sqrt{4 - x}}{2 \cdot - 1}$

Step 3.7.2

Multiply $2$ by $- 1$ .

$y = \frac{- 2 \pm 2 \sqrt{4 - x}}{- 2}$

Step 3.7.3

Simplify $\frac{- 2 \pm 2 \sqrt{4 - x}}{- 2}$ .

$y = 1 \pm \sqrt{4 - x}$

Step 3.7.4

Change the $\pm$ to $-$ .

$y = 1 - \sqrt{4 - x}$

Step 3.8

The final answer is the combination of both solutions.

$y = 1 + \sqrt{4 - x}$

$y = 1 - \sqrt{4 - x}$

$y = 1 + \sqrt{4 - x}$

$y = 1 - \sqrt{4 - x}$

Step 4

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = 1 + \sqrt{4 - x}, 1 - \sqrt{4 - x}$

Step 5

Verify if $f^{- 1} (x) = 1 + \sqrt{4 - x}, 1 - \sqrt{4 - x}$ is the inverse of $f (x) = - x^{2} + 2 x + 3$ .

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Step 5.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = - x^{2} + 2 x + 3$ and $f^{- 1} (x) = 1 + \sqrt{4 - x}, 1 - \sqrt{4 - x}$ and compare them.

Step 5.2

Find the range of $f (x) = - x^{2} + 2 x + 3$ .

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Step 5.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, 4]$

Step 5.3

Find the domain of $1 + \sqrt{4 - x}$ .

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Step 5.3.1

Set the radicand in $\sqrt{4 - x}$ greater than or equal to $0$ to find where the expression is defined.

$4 - x \geq 0$

Step 5.3.2

Solve for $x$ .

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Step 5.3.2.1

Subtract $4$ from both sides of the inequality.

$- x \geq - 4$

Step 5.3.2.2

Divide each term in $- x \geq - 4$ by $- 1$ and simplify.

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Step 5.3.2.2.1

Divide each term in $- x \geq - 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 4}{- 1}$

Step 5.3.2.2.2

Simplify the left side.

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Step 5.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 4}{- 1}$

Step 5.3.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 4}{- 1}$

Step 5.3.2.2.3

Simplify the right side.

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Step 5.3.2.2.3.1

Divide $- 4$ by $- 1$ .

$x \leq 4$

Step 5.3.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, 4]$

Step 5.4

Find the domain of $f (x) = - x^{2} + 2 x + 3$ .

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Step 5.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Step 5.5

Since the domain of $f^{- 1} (x) = 1 + \sqrt{4 - x}, 1 - \sqrt{4 - x}$ is the range of $f (x) = - x^{2} + 2 x + 3$ and the range of $f^{- 1} (x) = 1 + \sqrt{4 - x}, 1 - \sqrt{4 - x}$ is the domain of $f (x) = - x^{2} + 2 x + 3$ , then $f^{- 1} (x) = 1 + \sqrt{4 - x}, 1 - \sqrt{4 - x}$ is the inverse of $f (x) = - x^{2} + 2 x + 3$ .

$f^{- 1} (x) = 1 + \sqrt{4 - x}, 1 - \sqrt{4 - x}$

Step 6